Answer
$\frac{{dy}}{{dx}} = \frac{{ - 2{e^y} - {e^x}y}}{{2x{e^y} + {e^x}}}$
Work Step by Step
$$\eqalign{
& 2x{e^y} + y{e^x} = 3 \cr
& {\text{Differentiating both sides of the equation with respect to }}x \cr
& \underbrace {\frac{d}{{dx}}\left[ {2x{e^y}} \right]}_{{\text{use product rule}}} + \underbrace {\frac{d}{{dx}}\left[ {y{e^x}} \right]}_{{\text{use product rule}}} = \frac{d}{{dx}}\left[ 3 \right] \cr
& 2x\frac{d}{{dx}}\left[ {{e^y}} \right] + 2{e^y}\frac{d}{{dx}}\left[ x \right] + y\frac{d}{{dx}}\left[ {{e^x}} \right] + {e^x}\frac{d}{{dx}}\left[ y \right] = \frac{d}{{dx}}\left[ 3 \right] \cr
& y{\text{ is a function of }}x,{\text{ using the chain rule we have}} \cr
& 2x{e^y}\frac{{dy}}{{dx}} + 2{e^y}\left( 1 \right) + y\left( {{e^x}} \right) + {e^x}\frac{{dy}}{{dx}} = 0 \cr
& 2x{e^y}\frac{{dy}}{{dx}} + 2{e^y} + {e^x}y + {e^x}\frac{{dy}}{{dx}} = 0 \cr
& {\text{Collect the terms that contains }}\frac{{dy}}{{dx}} \cr
& 2x{e^y}\frac{{dy}}{{dx}} + {e^x}\frac{{dy}}{{dx}} = - 2{e^y} - {e^x}y \cr
& {\text{Solve this equation for }}\frac{{dy}}{{dx}} \cr
& \left( {2x{e^y} + {e^x}} \right)\frac{{dy}}{{dx}} = - 2{e^y} - {e^x}y \cr
& \frac{{dy}}{{dx}} = \frac{{ - 2{e^y} - {e^x}y}}{{2x{e^y} + {e^x}}} \cr} $$
